Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

50
ADVANCED ABSTRACT ALGEBRA
Z ∆2Z ∆4Z ∆8Z ∆16Z
∆ − − − − − − − −
We can not end to
Definition:
= (1).
Let G = G ∆G ∆G ∆ − − − − ∆ G r = ( ) be a composition series and suppose that
0 1 2 1
G = H ∆H ∆ − − − − ∆ H r = ( )
0 1 1
is another composition series of the same length r. We say that these series are equivalent if ∃ some
such that
G H
i
G
≅
– 1 σ ( i)–
1
i
H
σ ( i)
∀ i.
Example 3.
Let G = < x > , 0 ( G)
= 6
(from unit I).
2
3
Let G1
= < x >. and H1
= < x >
We have two composition series:
G ∆G ∆G
1 2 1
G ∆ H ∆H
= ( ) and
= ( )
1 2
1
These two series are equivalent, as
G G
1
H
≅
1
≅ Z
( 1) 2 and
NG
∴Z
σ
' '∆
G1
G Z
( 1) ≅ ≅ H
F
HG
Θ G G
1
1
3
x G G x G
= < > ,
2
x G
= 2, 1 H
= < > , = 3
3
< >
1 < x > H
and take σ = ( 12)
∈S2
Theorem 2:
Jorden-Holder Theorem:
This theorem asserts that, upto equivalence, a group has at most one composition series.
Statement:
1
Suppose that G is a group that has a composition series. Then any two composition series of G have the same
length and are equivalent.
Proof:
Let G = G > G > − − − − > G r = ( )
0 1 1
and G = H0 > H1 > − − − − > H s = ( 1)
be two composition series of G. We use induction on r, the length of one of the composition series.
I
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